Selina Concise Solutions for Chapter 14 Linear Equations in One Variable Class 8 ICSE Mathematics

Exercise 14A


Solve the following equations:

1. 20 = 6 + 2x
Solution
20 = 6 + 2x
⇒ 20 – 6 = 2x
⇒ 14 = 2x
⇒ 7 = x
⇒ x = 7

2. 15 + x = 5x + 3
Solution
15 – 3 = 5x – x
⇒ 12 = 4x
⇒ 3 = x
⇒ x = 3

3. (3x + 2)/(x – 6)  = -7
Solution
3x + 2 = -7 (x – 6) (by cross multiplying)
⇒ 3x + 2 = -7x + 42
⇒ 3x + 7x = 42 – 2
⇒ 10x = 40
⇒ x = 4

4. 3a – 4 = 2(4 – a)
Solution
3a – 4 = 8 – 2a
⇒ 3a + 2a = 8 + 4
⇒ 5a = 12
⇒ a = 2.4

5. 3(b – 4) = 2(4 – b)
Solution
⇒ 3b – 12 = 8 – 2b
⇒ 3b + 2b = 8 + 12
⇒ 5b = 20
⇒ b = 20/5
⇒ b = 4

6. (x + 2)/9 = (x + 4)/11
Solution
⇒ 11(x + 2) = 9(x + 4) (by cross multiplication)
⇒ 11x + 22 = 9x + 36
⇒ 11x – 9x = 36 – 22
⇒ 2x = 14
⇒ x = 14/2
⇒ x = 7

7. (x – 8)/5 = (x – 12)/9 
Solution
⇒ 9(x – 8) = 5(x – 12)   (by cross multiplication)
⇒ 9x - 72 = 5x – 60
⇒ 9x – 5x = -60 + 72
⇒ 4x = 12
⇒ x = 12/4
⇒ x = 3

8. 5(8x + 3) = 9(4x + 7) 

Solution 8: 
⇒ 40x + 15 = 36x + 63
⇒ 40x – 36x = 63 – 15
⇒ 4x = 48
⇒ x = 48/4
⇒ x = 12

9. 3(x + 1) = 12 + 4(x – 1)
Solution
3(x + 1) = 12 + 4(x – 1)
⇒ 3x + 3 = 12 + 4x – 4
⇒ 3x – 4x = 12 – 4 – 3
⇒ -x = 5
⇒ x = -5

10. 3x/4 - ¼ (x – 20) = x/4 + 32 
Solution
⇒ 3x/4 – x/4 + 5 = x/4 + 32
⇒ 3x/4 – x/4 – x/4 = 32 – 5
⇒ (3x – x – x)/4 = 27
⇒ x/4 = 27
⇒ x = 27 × 4
⇒ x  = 108

11. 3a – 1/5 = a/5 + 5 2/5 
Solution
⇒ 3a - a/5 = 5 2/5 + 1/5
⇒ 3a – a/5 = 27/5 + 1/5
⇒ 3a × 5 – a/5 × 5 = 27/5 × 5 + 1/5 × 5
(Multiplying each term by 5)
⇒ 15a – a = 27 + 1
⇒ 14a = 28
⇒ a = 28/14
⇒ a = 2

12. x/3 – 2 ½ = 4x/9  – 2x/3 
Solution
⇒ x/3 – 5/2 = 4x/9 – 2x/3
Since, L.C.M. of denominators 3, 2, 9  and 3 = 18
⇒ x/3 ×18 – 5/2 ×18 = 4x/9 ×18 – 2x/3 ×18
[Multiplying each term by 18]
⇒ 6x – 45 = 8x – 12x
⇒ 6x + 12x – 8x = 45
⇒ 18x – 8x = 45
⇒ 10x = 45
⇒ x = 45/10
⇒ x = 4.5

13. 4(y + 2)/5 = 7 + 5y/13 
Solution
⇒ (4y + 8)/5 = 7 + 5y/13
⇒ (4y + 8)/5 = (91 + 5y)/13   (by cross multiplication)
⇒ 13(4y + 8) = 5(91 + 5y)
⇒ 52y + 104 = 455 + 25y
⇒ 52y – 25y = 455 – 104
⇒ 27y  = 351
⇒ y = 351/27
⇒ y = 13

14. (a + 5)/6 – (a + 1)/9 = (a + 3)/4 
Solution
Since L.C.M. of denominators  6, 9 and 4 = 36
∴ (a + 5)/6 × 36  - (a + 1)/9 × 36 = (a + 3)/4  × 36
(Multiplying each term by 36)
⇒ 6(a + 5) – 4(a + 1) = 9(a + 3)
⇒ 6a + 30 – 4a – 4 = 9a + 27
⇒ 6a -  4a – 9a = 27 – 30 + 4
⇒ 6a – 13a = 1
⇒ -7a = 1
⇒ a = - 1/7

15. (2x – 13)/5 – (x – 3)/11 = (x – 9)/5 + 1 
Solution
⇒ (2x – 13)/5 – (x – 3)/11 = (x – 9)/5 + 1/1
Since, L.C.M. of denominators 5,11, 5 and 1 = 55
∴ (2x – 13)/5 × 55 – (x – 3)/11 ×55 = (x – 9)/5 × 55 + 1/1 × 55
⇒ 11(2x – 13) – 5(x – 3) = 11(x – 9) + 55
⇒ 22x – 143 – 5x + 15 = 11x – 99 + 55
⇒ 22x – 5x – 11x = - 99 + 55 + 143 – 15
⇒ 6x = 198 - 114
⇒ 6x = 84
⇒ x = 84/6
⇒ x = 14

16. 6(6x – 5) – 5(7x – 8) = 12(4 – x) = 1
Solution
6(6x – 5) – 5(7x – 8) = 12(4 – x) + 1
⇒ 36x - 30 – 35x + 40 = 48 – 12x + 1
⇒ x + 12x = 49 – 10
⇒ 13x = 39
⇒ x = 19/13
⇒ x = 3

17. (x – 5)(x + 3) = (x – 7)(x + 4) 
Solution
(x – 5)(x + 3) = (x – 7)(x + 4)
⇒ x2 + 3x – 5x – 15 = x2 + 4x – 7x – 28
⇒ -2x – 15 = -3x – 28
⇒ 3x – 2x = 15 – 28
⇒ x = - 13

18. (x – 5)2 – (x + 2)2 = -2
Solution
(x – 5)2 – (x + 2)2 = -2
⇒ (x2 – 10x + 25) – (x2 + 4x + 4) = -2
⇒ x2 – 10x + 25 – x2 – 4x – 4 = -2
⇒ -10x – 4x + 25 – 4 = -2
⇒ -14x = 4 – 2 – 25 = - 23
⇒ x = -23/-14
= 23/24
= 1 9/14 

19. (x – 1)(x + 6) – (x – 2)(x – 3) = 3 
Solution
(x – 1)(x + 6) – (x – 2)(x – 3) = 3
⇒ x2 – x + 6x – 6 – (x2 – 3x – 2x + 6) = 3
⇒ x2 – x + 6x – 6 – x2 + 3x + 2x – 6 = 3
⇒ -x + 6x + 3x + 2x – 6 – 6 = 3
⇒ -x + 11x – 6 – 6 = 3
⇒ 10x = 15
⇒ x = 15/10 = 3/2
⇒ x = 1 ½

20. 3x/(x + 6) – x(x + 5) = 2
Solution
3x/(x + 6) – x/(x + 5) = 2
⇒ {3x(x + 5) – x(x + 6)}/(x + 6)(x + 5) = 2
⇒ (3x2 + 15x – x2 – 6x)/(x2 + 5x + 6x + 30) = 2
⇒ (2x2 + 9x)/(x2 + 11x + 30) = 2
⇒ 2x2 + 9x = 2(x2 + 11x + 30)
⇒ 2x2 + 9x = 2x2 + 22x + 60
⇒ 2x22 – 2x2 + 9x – 22x = 60
⇒ -13x = 60
⇒ x = - 60/13
⇒ x = -4 8/13

21. 1/(x – 1) + 2/(x – 2) = 3/(x – 3) 
Solution 
{1(x – 2) + 2(x – 1)}/{(x – 1)(x – 2)} = 3/(x – 3)
⇒ (x – 2 + 2x - 2)/(x2 – 2x – x + 2) = 3/(x – 3)
⇒ (3x – 4)/(x2 – 3x + 2) = 3/(x – 3)
⇒ (x – 3)(3x – 4) = 3(x2 – 3x + 2)
⇒ 3x2 – 4x – 9x + 12 = 3x2 – 9x + 6
⇒ 3x2 – 13x – 3x2 + 9x = 6 – 12
⇒ - 4x = -6
⇒ x = -6/-4 = 3/2
⇒ x = 1 ½

22. (x – 1)/(7x – 14) = (x – 3)/(7x – 26) 
Solution
(x – 1)/(7x – 14) = (x – 3)/(7x – 26)
⇒ (x – 1)(7 – 26) = (7x – 14)(x – 3)
⇒ 7x2 – 7x – 26x + 26 = 7x2 – 14x – 21x + 42
⇒ -33x + 26 = -35x + 42
⇒ 35x – 33x   = 42 – 26
⇒ 2x = 16
⇒ x = 8

23. 1/(x – 1) – 1/x = 1/(x + 3) – 1/(x + 4) 
Solution
1/(x – 1) – 1/x = 1/(x + 3) – 1/(x + 4)
⇒ {x – (x – 1)}/(x – 1)x = {(x + 4) – (x + 3)}/{(x + 3)(x + 4)}
= 1/(x – 1)x = 1/(x + 3)(x + 4)
= (x + 3)(x + 4)
= x(x – 1)
⇒ x2+ 4x  + 3x + 12 = x2 – x
⇒ x2 + 7x – x2 + x = -12
⇒ 8x = -12
⇒ x = -12/8 = -3/2
⇒ x = -1 ½

24. Solve : 2x/3 – (x – 1)/6 + (7x – 1)/4  = 2 1/6 
Solution
2x/3 – (x – 1)/6 + (7x – 1)/4 = 2 1/6
⇒ 2x/3 – (x – 1)/6 + (7x – 1)/4 = 13/6
⇒ (8x – 2x + 2 + 21x – 3)/12 = 26 (L.C.M. of 3, 6, 4, 6 = 12)
⇒ 27x – 1 = 26
⇒ 27x = 26 + 1
⇒ x = 27/27 = 1
Now, 1/a + 5x = 8
⇒ 1/a + 5 × 1 = 8
⇒ 1/a + 5 = 8
⇒ 1/a = 8 – 5 = 3
∵ 3a = 1
⇒ a = 1/3
∴ x = 1 and a = 1/3


25. Solve: (4 – 3x)/5 + (7 – x)/3 + 4 1/3  = 0 
Hence, find the value of ‘p’ if 2p – 2x + 1 = 0 
Solution
⇒ (4 – 3x)/5 + (7 – x)/3 + 4 1/3 = 0
⇒ (4 – 3x)/5 + (7 – x)/3 + 13/3 = 0
⇒ (12 – 9x + 35 – 5x + 65)/15 = 0
(L.C.M of 5, 3, 3 = 15)
-14x + 112 = 0
⇒ -14x = -112
⇒ x = -112/-14
⇒ x = 8 
Hence x = 8
Now, 3p – 2x + 1 = 0
⇒ 3p – 2 × 8 + 1 = 0
⇒ 3p – 16 + 1 = 0
⇒ 3p  - 15 = 0
⇒ 3p = 15
⇒ p = 5 

26. Solve : 0.25 + 1.95/x = 0.9 
Solution
0.25 + 1.95/x = 0.9
⇒ 0.25x + 1.95 = 0.9x
⇒ 0.9x – 0.25x = 1.95
⇒ 0.65x = 1.95
⇒ x = 1.95/0.65 = 3
Hence, x = 3

27. Solve : 5x – {4x + (5x – 4)/7} = (4x – 14)/3
Solution
5x – (4x + (5x – 4)/7) = (4x – 14)/3
⇒ 5x – (28x + 5x – 4)/7 = (4x – 14)/3
⇒ (35x – 33x  + 4)/7 = (4x  - 14)/3
⇒ 3 × (2x + 4) = 7 × (4x – 14)
⇒ 6x + 12 = 28x – 98
⇒ 22x = 98 + 12
⇒ x = 110/22 = 5

Exercise 14 B


1. Fifteen less than 4 times a number is 9. Find the number.
Solution
Let the required number be x
4 times the number = 4x
15 less than 4 times the number = 4x - 15
According to the statement :
4x – 15 = 9
⇒ 4x = 9 + 15
⇒ 4x = 24
⇒ x = 6

2. If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Solution
Let Megha’s age = x years
Three times Megha’s age = 3x years
According to the statement :
x + 3x = 60
⇒ 4x = 60
⇒ x = 15
Megha’s age = 15 years

3. 28 is 12 less than 4 times a number. Find the number.
Solution
Let the required number be x
4 times the number = 4x
12 less than 4 times the number = 4x – 12
According to the statement
4x – 12 = 28
⇒ 4x = 28 + 12
⇒ 4x = 40
⇒ x = 10
Required number = 10

4. Five less than 3 times a number is -20. Find the number.
Solution
Let the required number = x
3 times the number = 3x
5 less than 3 times the number = 3x – 5
According to statement :
3x – 5 = -20
⇒ 3x = -20 + 5
⇒ 3x = -15
⇒ x = -5
Required number = -5

5. Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she?
Solution
Let Neetu’s age = x years
3 times Neetu’s age = 3x years
Fifteen more than 3 times Neetu’s age = (3x + 15) years
4 times Neetu’s age = 4x
According to the statement :
4x = 3x + 15
⇒ 4x – 3x = 15
⇒ x = 15
Neetu’s age = 15 years

6. A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number
Solution
Let the required number = x
The number decreased by 30 = x – 30
14 decreased by 3 times the number = 14 – 3x
According to the statement :
x – 30 = 14 – 3x
⇒ x + 3x = 14 + 30
⇒ 4x = 44
⇒ x = 11
Required number = 11

7. A’s salary is same as 4 times B’s salary. If together they earn Rs. 3,750 a month, find the salary of each.
Solution
Let B’s salary = Rs. x
A’s salary = Rs. 4x
According to the statement :
x + 4x = 3750
⇒ 5x = 3750
⇒ x = 750
⇒ 4x = 750 × 4 = 3000
A’s salary = Rs. 3000
B’s salary = Rs. 750

8. Separate 178 into two parts so that the first part is 8 less than twice the second part.
Solution
Let first part = x
Second part = 178 – x
According to the problem :
First Part = 8 less than twice the second part
x = 2(178 – x) – 8
⇒ x = 356 – 2x – 8
⇒ x + 2x = 356 – 8
⇒ 3x = 348
⇒ x = 116
First Part = 116
Second Part = 178 – x = 178 – 116 = 62
Alternative Method :
Let Second part = x
First part = 2x – 8
According to the problem :
x + 2x – 8 = 178
⇒ x + 2x = 178 + 8
⇒ 3x = 186
⇒ x = 62
First part = 2x – 8 = 2×62 – 8 = 124 – 8 = 116
First part = 116
Second part = 62

9. Six more than one-fourth of a number is two-fifth of the number. Find the number.
Solution
Let the required number = x
∴ One-fourth of the number = x/4
Two –fifth of the number = 2x/5
According to the statement :
2x/5 = 6 + x/4
⇒ 2x/5 – x/4 = 6/1
⇒ 2x/5 ×20 – x/4 × 20 = 6×20
[Multiplying each term by 20 because L.C.M. of 5,4 and 1 is 20]
⇒ 8 – 5x = 120
⇒ 3x = 120
⇒ x = 120/3
⇒ x = 40
Required number = 40

10. The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.
Solution
Let width of a rectangle = x cm
Length of the rectangle = 2x cm
Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm
Given perimeter = 54 cm
6x = 54
⇒ x = 9
Length = 2x = 2 x 9 = 18 cm

11. A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the original rectangle.
Solution
Let width of the original rectangle = x cm
Length of the original rectangle = (2x – 5)cm
Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm
New width of the rectangle = (x + 2) cm
New perimeter = 2[Length + Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm
Given; new perimeter = 74 cm
6x – 16 = 74
⇒ 6x = 74 + 16
⇒ 6x = 90
⇒ x = 15
Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm
Width of the original rectangle = x = 15 cm

12. The sum of three consecutive odd numbers is 57. Find the numbers.
Solution
Let the three consecutive odd numbers be x, x + 2, x + 4.
According to the statement :
x + x + 2 + x + 4 = 57
⇒ x + x + x = 57 – 2 – 4
⇒ 3x = 51
⇒ x = 17
Three consecutive odd numbers are 17, 19, 21

13. A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solution
Let present age of the son = x years
present age of the man = 3x years
In 12 years :
Son’s age will be = (x + 12) years
The man’s age will be = (3x + 12) years
According to the statement :
3x + 12 = 2(x + 12)
⇒ 3x + 12 = 2x + 24
⇒ 3x – 2x = 24 – 12
⇒  x = 12
3x = 3 x 12 = 36
Hence, present age of the man = 36 years
Present age of the son = 12 years.

14. A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Solution
Man’s age = 42 years
Son’s age = 12 years
Let after x years the age of the son will be half the age of the man.
Man’s age after x years = (42 + x) years
Son’s age after x years = (12 + x) years
According to the statement :
12 + x = (42 + x)/2
⇒ 2(12 + x) = 42 + x
(by cross multiplying)
⇒ 24 + 2x = 42 + x
⇒ 2x – x = 42 – 24
⇒ x = 18
Hence after 18 years, the age of the son will be half the age of the man.

15. A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Solution
Total distance of the trip = 136km.
Let part of trip covered at 18 km/hr. = x km.
∴ Distance of the trip covered at 15 km/hr = (136 – x)km
Time taken by the man to cover x km = Distance/Speed = x/18  hours
Time taken by the man to cover (136 – x) km = (136 – x)/15 hours
Time time taken by the man to cover (136 – x) km = (136 – x)/15 hours
Total time taken by the man to cover a trip of 136 km = 8 hours
∴ x/18 + (136 – x)/15 = 8
⇒ x/18 ×90 + (136 – x)/15 ×90  = 8 × 90
[Multiplying each term by 90 because L.C.M.  of denominators = 90]
⇒ 5x + 6 (136 – x)  = 720
⇒ 5x + 816 – 6x = 720
⇒  5x – 6x = 720 – 816
⇒ -x = -96
⇒ x = 96
∴ Part of the trip covered at 18km/hr = 96 km

16. The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Solution
Let one number = x
Second number = x + 3 [Difference of two numbers is 3]
According to the statement :
(x + 3)2 – (x)2 = 69
⇒ (x)2 + (3)2 + 2 × x × 3 – x2 = 69
⇒ x2 + 9 + 6x – x2 = 69
⇒ 6x = 69 – 9
⇒ 6x = 60
⇒ x = 60/6
⇒ x = 10
One number = 10
Second number = x + 3 = 10 + 3 = 13

17. Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.
Solution
Let two consecutive natural numbers = x, x + 1
∴ One –fourth of the smaller = x/4
One-fifth of the greater = (x + 1)/5
According to the statement :
x/4 =  (x + 1)/5 + 1
⇒ x/4 – (x + 1)/5 = 1
⇒ (5x – 4(x + 1))/20 = 1
⇒ (5x - 4x – 4)/20 = 1
⇒ (x – 4)/20 = 1
⇒ x – 4 = 20  (Cross-multiplying)
⇒ x = 20 + 4
⇒ x = 24
∴ x + 1 = 24 + 1
= 25
Two consecutive numbers are 24 and 25

18. Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.
Solution
Let the three consecutive whole numbers be x, x + 1 and x + 2
According to the statement:
x/5 + (x + 1)/3 + (x + 2)/4 = 40
⇒ x/5 × 60 + (x + 1)/3 × 60 + (x + 2)/4 × 60 = 40 × 60
{Multiplying each term by 60 because L.C.M. of denominators = 60]
⇒ 12x + 20(x + 1) + 15(x + 2) = 2400
⇒ 12x + 20x + 20 + 15x + 30 = 2400
⇒ 12x + 20x + 15x = 2400 – 20 – 30
⇒ 47x = 2350
⇒ x = 2350/47
⇒ x = 50
x + 1 = 50 + 1 = 51
x + 2 = 50 + 2 = 52
Three consecutive whole numbers are 50, 51 and 52

19. If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Solution
Let x be added to each number, then the numbers will be 5 + x, 11 + x, 15 + x and 31 + x
According to the condition
(5 + x)/(11 + x) = (15 + x)/(31 + x)
By cross multiplication,
(5 + x)(31 + x) = (15 + x)(11 + x)
⇒ 155 + 5x + 31x + x2  = 165 + 11x + 15x + x2
⇒ 155 + 36x + x2 = 165 + 26x + x2
⇒ 36x + x2 – 26x – x2 = 165 – 155
⇒ 10x = 10
⇒ x = 10/10
⇒ x = 1
1 should be added

20. The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7: 4. Find their present ages.
Solution
Let present age of son = x year
Then age of his father = 2x
8 years hence,
Age of son = (x + 8) years and age of father = (2x + 8) years
According to the condition,
(2x + 8)/(x + 8) = 7/4
⇒ 8x + 32 = 7x + 56
⇒ 8x – 7x = 56 – 32
⇒ x = 24
Present age of son = 24 years
and age of father = 2x = 2 x 24 = 48 years
Hence age of man = 48 years and age of his son = 24 year

Exercise 14 C


1. Solve:  
(i) 1/3x – 6 = 5/2 
(ii) 2x/3 – 3x/8 = 7/12 
(iii) (x + 2)(x + 3) + (x - 3)(x – 2) – 2x(x + 1) = 0 
(iv) 1/10 – 7/x = 35 
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0 
(vi) x + 7 – 8x/3 = 17x/6 – 5x/8 
(vii) (3x – 2)/4 – (2x + 3)/3 = 2/3 – x
(viii) (x + 2)/6  – ((11 - x)/3 – ¼) = (3x – 4)/12 
(ix) 2/5x – 5/3x = 1/15 
(x) (x + 2)/3 – (x + 1)/5 = (x – 3)/4 – 1
(xi) (3x – 2)/3 + (2x + 3)/2 = x + 7/6 
(xii) x - (x - 1)/2 = 1 - (x - 2)/3
(xiii) (9x + 7)/2 – (x – (x – 2)/7) = 36 
(xiv) (6x + 1)/2 + 1 = (7x – 3)/3 
Solution
(i) 1/3x – 6 = 5/2
⇒ 1/3x = 5/2 + 6/1
⇒ 1/3x = (5 × 1)/(2 × 1) + (6 × 2)/(1 × 2)
⇒ 1/3x = 5/2 + 12/2
⇒ 1/3x = (5 + 12)/2
⇒ 1/3x  = 17/2
⇒ x = (17 × 3)/(2 ×1) = 51/2 = 25 ½

(ii) 2x/3 – 3x/8 = 7/12
L.C.M. of 3 and 8 = 2 × 2 × 2 × 3 = 24
∴ (2x × 8)/(3 × 8) – (3x × 3)/(8 × 3) = 7/12
⇒ 16x/24 – 9x/24 = 7/12
⇒ 16x/24 – 9x/24 = 7/12
⇒ (16x – 9x)/24 = 7/12
⇒ 7x/24 = 7/12
⇒ x = (7 × 24)/(12 × 7) = 2
∴ x = 2

(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
⇒ [x2 + (2 + 3)x + 2 × 3] + [x2 + (-3 – 2)x + (-3)(-2)] – 2x2 – 2x = 0
⇒ x2 + 5x + 6 + x2 – 5x + 6 – 2x2 – 2x = 0
⇒ x2 + x2 – 2x2 + 5x – 5x – 2x + 6 + 6 = 0
⇒ -2x + 12 = 0
Subtracting 12 from both sides,
-2x + 12 – 12 = 0 – 12
⇒ -2x = -12
Dividing by – 2
-2x/-2 = -12/-2
⇒ x = 6
∴ x = 6
Verification
L.H.S. = (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1)
= (6 + 2)(6 + 3) + (6 – 3)(6 – 2) – 2 × 6(6 + 1)
= 8 × 9 + 3 × 4 – 12 × 7
= 72 + 12 – 84
= 84 – 84
= 0 = R.H.S.

(iv) 1/10 – 7/x = 35
⇒ -7/x = 35 – 1/10
⇒ -7/x = (35 × 10)/(1 × 10) – (1 × 1)/(10 × 1)
⇒ -7/x = (350 – 1)/10
⇒ 1/x = (350 – 1)/(10 × (-7))
⇒ x = 349/(-70) = (-70)/349

(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
⇒ 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
⇒ 13x – 52 – 3x + 27 – 5x – 20 = 0
⇒ 13x – 3x – 5x – 52 + 27 – 20 = 0
⇒ 13x – 8x – 72 + 27 = 0
⇒ 5x – 45 = 0
Dividing by 5,
5x/5 – 45/5 = 0
⇒ x – 9 = 0
⇒ x = 9
Verification,
L.H.S. = 13(x – 4) – 3(x – 9) – 5(x + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 × 5 – 3 × 0 – 5 × 13
= 65 – 0 – 65
= 0
= R.H.S.

(vi) x + 7 – 8x/3 = 17x/6 – 5x/8
⇒ (3(x + 7) – 8x)/3  = (68x – 15x)/24
⇒ (-5x + 21)/3 = 53x/24
⇒ 3 × 53x = 24(-5x + 21)
⇒ 159x = -120x + 504
⇒ 159 x  + 120x  = 504
⇒ 279x = 504
⇒ x = 504/279
⇒ 168/93 = 56/61
∴ x = 1 25/31

(vii) (3x – 2)/4 – (2x + 3)/3 = 2/3 – x
= (3x – 2)/4 – (2x + 3)/3 = 2/3 – x/1
⇒ (3(3x – 2) – 4(2x + 3))/12 = (2 × 1)/(3 × 1) – (x × 3)/(1 × 3)
⇒ (9x – 6 – 8x – 12)/12 = (2  - 3x)/3
⇒ (x – 18)/12 = (2 – 3x)/3
⇒ 3(x – 18) = 12(2 – 3x)
⇒ 3x – 54 = 24 – 36x
⇒ 3x + 36x = 24 + 54
⇒ 39x = 78
x  = 78/39 = 2
∴ x = 2

(viii) (x  + 2)/6 – ((11 – x)/3 – ¼) = (3x – 4)/12
⇒ (x + 2)/6 – (4(11 - x) – 1 × 3)/12) = (3x – 4)/12
⇒ (x + 2)/6 – (44 + 4x + 3)/12 = (3x – 4)/12
⇒ (2(x + 2) – 41 + 4x)/12 = (3x – 4)/12
⇒ (2x + 4 – 41 + 4x)/12 = (3x – 4)/12
⇒ (6x – 37)/12 = (3x – 4)/12
⇒ 12(6x – 37) = 12(3x – 4)
⇒ 72x – 444 = 36x – 48
⇒ 72x – 36x = -48 + 444
⇒ 36x = 396
⇒ x = 396/36 = 11
∴ x = 11

(ix)  2/5x – 5/3x = 1/15
⇒ (2 × 3)/(5x × 3) – (5 × 5)/(3x × 5) = 1/15
⇒ (6 – 25)/15x  = 1/15
⇒ -19/15x = 1/15
⇒ -19/x = 15/15
⇒ -19 = x
∴ x = - 19

(x) (x + 2)/3 – (x + 1)/5 = (x – 3)/4 – 1
(L.C.M. of 3 and 5 = 15)
⇒ (5(x + 2) – 3(x + 1))/15  = (x – 3 – 4)/4
⇒  (5x + 10 – 3x – 3)/15 = (x – 7)/4
⇒ (2x + 7)/15 = (x – 7)/4
⇒ 4(2x + 7) = 15(x – 7)
⇒ 8x + 28 = 15x – 105
⇒ 8x – 15x = -105 – 28
⇒ -7x = -33
x = -133/-7
∴ x = 19

(xi) (3x – 2)/3 + (2x + 3)/2 = x + 7/6
⇒ (2(3x - 2) + 3(2x + 3))/6  = x + 7/6 
⇒ (6x – 4 + 6x + 9)/6 = (6x + 7)/6
⇒ (12x + 5)/6 = (6x + 7)/6
⇒ 6(12x + 5) = 6(6x + 7)
⇒ 72x – 36x = 42 – 30
⇒ 36x = 12
x = 12/36
∴ x = 1/3

(xii) x – (x – 1)/2 = 1 – (x – 2)/3
⇒ (2(x) – 1(x  - 1))/2 = (3(1) – 1(x - 2))/3
⇒ (2x – x + 1)/2 = (3 – x + 2)/3
⇒ (1x + 1)/2  = (5 – x)/3
⇒ 3(x  + 1) = 2(5 – x)
⇒ 3x + 3 = 10 – 2x
⇒ 3x + 2x  = 10 – 3
⇒ 5x = 7
∴ x  = 7/5

(xiii) (9x + 7)/2 – (x – (x – 2)/7) = 36
⇒ (9x + 7)/2 – (7 × x – 1(x – 2))/7 = 36
⇒ (9x + 7)/2 – ((7x – x – 2)/7) = 36
⇒ (9x + 7)/2 – ((6x – 2)/7) = 36
⇒ (7(9x + 7) + 2(-6x + 2))/14 = 36
⇒ (63x + 49 – 12x + 4)/14 = 36
⇒ (51x + 53)/14 = 36
⇒ 51x + 53 = 14 × 36
⇒ 51x = 504 – 53 
⇒ 51x = 459
⇒ x = 459 /51
∴ x = 9

(xiv) ((6x + 1)/2) + 1 = (7x – 3)/ 3
⇒ ((6x + 1) + 1 × 2)/2 = (7x – 3)/3
⇒ (6x + 1 + 2)/2 = (7x – 3)/3
⇒ (6x + 3)/2 = (7x – 3)/3
⇒ 3(6x + 3) = 2(7x – 3)
⇒ 18x + 9 = 14x – 6
⇒ 18x – 14x = -6 – 9
⇒ 4x = -15
∴ x = -15/4

2.  After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.
Solution
Let present age = x years
According to question,
(x + 12) = 3(x – 4)
x + 12 = 3x – 12
2x = 24
⇒ x = 12 years
Present age = 12 years

3. A man sold an article for 7396 and gained 10% on it. Find the cost price of the article
Solution
S.P. of article = ₹ 396
Gain = 10%
Let cost price = ₹ x
∴ S.P. = (x × (100 + 10))/100 = 110/100x
∴ 110/100x = 396
⇒ x = (396 × 100)/110 = 360
Cost price of an article = ₹ 360

4. The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Solution
Let the first number = x
and the second number = y
According to question,
x + y = 4500 …(i)
and 10% x = 12.5% y
i.e. 10x = 12.5y
⇒ x = 12.5/10y  …(ii)
Substitute the value of x in equation (i),
12.5/10y + y = 45,000
⇒ 12.5y + 10y = 45,000
⇒ 22.5y = 45,000
y = 45,000/22.5 = 2000
Now, put the value of y in equation (ii)
x = 12.5/10 × 2000
⇒ x = 2500
Hence, the numbers are 2500 and 2000

5. The sum of two numbers is 405 and their ratio is 8:7. Find the numbers.
Solution
Let the first number = x
and the second number = 7
According to the question, x + y = 405 ...(i)
and the numbers are in the ratio 8 : 7
i.e  8x/7y = 1
⇒ 8x = 7y
⇒ x = 7/8 y
Now, substitute the value of x in equation (i)
7/8y + y = 405
⇒ 7y + 8y = 450 × 8
⇒ 15y = 3240
⇒ y = 3240/15
⇒ y = 216
Now, put the value of y in equation(ii)
x = 7/8 × 216
⇒ x  = 189
Hence, the numbers are 189 and 216.

6. The ages of A and B are in the ratio 7:5. Ten years hence, the ratio of their ages will be 9:7. Find their present ages.
Solution
Ratio in the present ages of A and B = 7 : 5
Let age of A = 7x years
Let age of B = 5x years
10 years hence,
Then age of A = 7x + 10 years
and age of B = 5x + 10 years
According to the condition,
(7x + 10)/(5x + 10) = 9/7
By crossing multiplication
7(7x + 10) = 9(5x + 10)
⇒ 49x + 70 = 45x + 90
⇒ 49x – 45x = 90 – 70
⇒ 4x = 20
⇒ x = 5
Present age of A = 7x = 7×5 = 35 years
and present age of B = 5x = 5×5 = 25 years

7. Find the number whose double is 45 greater than its half.
Solution
Let the required number = x
Double of it = 2x
And half of it = x/2
According to condition,
2x – x/2 = 45
⇒ (4x – x)/2 = 45
⇒ 3/2x = 45
⇒ x = (45 × 2)/3 = 30
Required number = 30

8. The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution
Let first number = x
and The second number = x + 1
According to the condition,
∴ (x + 1)2 – (x)2 = 31
⇒ x2 + 2x + 1 – x2 = 31
⇒ 2x = 31 – 1 = 30
⇒ x = 30/2 = 15
First number = 15
and second number = 15 + 1 = 16
Hence, the numbers are 15, 16

9. Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution
Let the required number = x
5 times of it = 5x
Twice of it = 2x
According to the condition,
5x – 5 = 2x + 4
⇒ 5x – 2x = 4 + 5
⇒ 3x = 9
⇒ x = 3
Required number = 3

10. The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes 2/3. Find the original fraction.
Solution
Let denominator of the original fraction = x
Then numerator = x – 5
And fraction = (x  - 5)/x
According to the condition,
(x – 5 + 3)/(x + 3) = 4/5
⇒ (x – 2)/(x + 3) = 4/5
⇒ 5(x – 2) = 4x + 12
(By cross multiplication)
⇒ 5x – 10 = 4x + 12
⇒ x = 22
∴ Original fraction = (x – 5)/x
= (22 – 5)/22
= 17/22
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